Thermal Conductivity Calculator
What conductivity is inferred from steady heat flow through a flat layer? Enter the stated inputs to calculate this model.
Result for the values shown.
Calculations run in your browser.
What to enter
| Input | Meaning and units |
|---|---|
| Heat flow rate (W) | Enter heat flow rate in W; use the same basis as the other measurements. |
| Layer thickness (m) | Enter layer thickness in m; use the same basis as the other measurements. |
| Area (m²) | Area in square metres. Convert both dimensions consistently before multiplying them. |
| Temperature difference magnitude (K) | Enter temperature difference magnitude in K; use the same basis as the other measurements. |
Understanding your result
Read Thermal conductivity (W/(m·K)) in the units and model stated on this page. The formula k=power*length/(area*delta) defines the relationship; measurement accuracy and model applicability are separate from numerical precision.
Common mistakes
- Mixing input units or changing the physical meaning of a variable without changing the model.
Check your calculation
- Use the worked example as a known reference case, then change one input at a time and check the direction and units of the response.
Calculation checks, sources and review limits
What conductivity is inferred from steady heat flow through a flat layer?
What conductivity is inferred from steady heat flow through a flat layer? Enter the stated inputs to calculate this model.
Common uses
- What conductivity is inferred from steady heat flow through a flat layer
How it works
k=power*length/(area*delta)
Worked example
Enter Heat flow rate: 20 W; Layer thickness: 0.1 m; Area: 2 m²; Temperature difference magnitude: 10 K. Results: Thermal conductivity: 0.1 W/(m·K).